Question

Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is

$$A.{\text{ 1:3}} \\\ B.{\text{ 3:2}} \\\ C.{\text{ 3:4}} \\\ D.{\text{ 1:2}} \\\$$

Answer 1

Hint- In order to solve the problem, first assume the speed of trains in some unknown variables. Then use the problem statement and the formula of the speed in terms of distance and time to form equations. Finally solve the equation to find out the relation between the speeds of trains.

Complete step-by-step answer:

Let the speeds of the trains be x m/s and y m/s respectively.
As the first train has speed x m/s and crosses the man standing in 27 seconds so the length of the train will be given by distance formula.

$$\because {\text{distance}} = {\text{speed}} \times {\text{time}} \\\ {\text{length of train}} = {\text{speed}} \times {\text{time}} \\\ {\text{length of train}} = x{\text{ }}m/s \times 27s \\\ {\text{length of train}} = 27x{\text{ }}m \\\$$

Also the second train has speed y m/s and crosses the man standing in 17 seconds so the length of the train will be given by distance formula.

$$\because {\text{distance}} = {\text{speed}} \times {\text{time}} \\\ {\text{length of train}} = {\text{speed}} \times {\text{time}} \\\ {\text{length of train}} = y{\text{ }}m/s \times 17s \\\ {\text{length of train}} = 17y{\text{ }}m \\\$$

As given in the problem the trains cross each other in 23 seconds.
So they must cover the length of both the trains in 23 seconds. As we know the length of both the trains. So let us use the distance formula.
Distance covered = sum of length of both the trains $= \left( {27x + 17y} \right)m$
As we know that both the trains are crossing each other so relative speed will be the sum of the speeds.
Speed $= \left( {x + y} \right)m/s$
Time taken = 23 s
As we know that

$$\because {\text{distance}} = {\text{speed}} \times {\text{time}} \\\ \therefore {\text{time}} = \dfrac{{{\text{distance}}}}{{{\text{speed}}}} \\\$$

So, let us substitute the value in order to find the relation between both the speeds of train
$\because {\text{time}} = \dfrac{{{\text{distance}}}}{{{\text{speed}}}} \\\ \Rightarrow 23s = \dfrac{{\left( {27x + 17y} \right)m}}{{\left( {x + y} \right)m/s}} \\\ \Rightarrow 23 = \dfrac{{27x + 17y}}{{x + y}} \\\$
Now let us solve the above equation by cross multiplying:

$$\Rightarrow 23\left( {x + y} \right) = 27x + 17y \\\ \Rightarrow 23x + 23y = 27x + 17y \\\ \Rightarrow 23y - 17y = 27x - 23x \\\ \Rightarrow 6y = 4x \\\ {\text{or}} \\\ \Rightarrow 4x = 6y \\\$$

Bringing the equation in $\dfrac{x}{y}$ in order to find the ration between the speeds of trains.

$$\Rightarrow \dfrac{x}{y} = \dfrac{6}{4} \\\ \Rightarrow \dfrac{x}{y} = \dfrac{3}{2} \\\$$

This is the ratio of speeds.
Hence, the ratio of speeds of both the trains is $3:2$ .
So, option B is the correct option.

Note- In order to solve such types of problems students must practically think of modifying the formula of speed and where to use. Also students must remember that the distance travelled by train in crossing a static pole or a man the distance covered by the train is the same as the length of train and in order to cross a platform or another train the distance travelled by the train is the sum of the length of train and platform/ other train.